Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ = / vo
vo = / sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
Answer:
4 m/s in negative acceleration
Explanation:
Acceleration = V- U/t
Where V is the final velocity
U is the initial velocity and t is the time given.
U = 65 m/s
V= 25 m/s
T= 10 seconds
Acceleration= (25m/s - 65m/s)÷10secs
= - 40/10
= -4m/s^2
Hence, it has a negative acceleration.
Answer:
200 N
Explanation:
For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.
The radially inward acceleration, or the centripetal acceleration is given by :
a = v² / r
where v is the speed at which the body is moving and r is the radius of the circle
Given-
m = 55kg
v = 14.1 m/s
r= 55m
We know that F = ma
⇒ F = m ( v²/ r )
⇒ F = 55 x 14.1 x 14.1 / 55
⇒ F =14.1 x 14.1 = 200 N
∴ <em>The force acting is 200 N</em>.
Answer:
Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
v2f=v2i−2⋅a⋅d
Isolate d on one side of the equation and solve by plugging your values
d=v2i−v2f2a
d=(15.02−10.02)m2s−22⋅2.0ms−2
d=31.3 m
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
vf=vi−a⋅t, which will get you
t=vi−vfa
t=(15.0−10.0)ms2.0ms2=2.5 s
Answer:
B) Pressure on the scale, not registered as weight.
Explanation:
This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point
